Question

does anyone know how to do piecewise functions
?

Answer 1

Go ahead and post a problem. Eventually someone with the talent will probably be able to walk you through the solution. You may have to be patient though, many of the East Coast/UK/Indian posters will have logged off by now.

Answer 2

What I'm saying is - keep checking back.

Answer 3

ok thank much for the help

Answer 4

f(X)=(2/3)x-3 if x less than or equal to 3
3x-10 if x is greater than 3

Answer 5

Do you need help graphing this?

Answer 6

yes

Answer 7

OK, it appears that you'll be graphing 2 rays. I will follow the slope-intercept equation y=(2/3)x - 3. But you'll only graph the part of the line that starts (above or below) 3 on the x-axis. This ray will proceed to the left, it needs a SOLID end-point.
Do you follow?

Answer 8

no i dont so u want me to graph y=(2/3)x - 3?

Answer 9

The other ray (not a full line) will follow the line y=3x-10, with a OPEN endpoint above or below the 3 on the x-axis - and proceeds to the right.

Answer 10

Sorry, finishing a comment.
Yes can you graph y=(2/3)x - 3 and y=3x-10 if you had too.
(otherwise I'll go over point - intercept graphing with you.)

Answer 11

i can individually

Answer 12

Great. Look at the second graph.
http://www.wolframalpha.com/input/?i=y%3D3x-10+and+y%3D%282%2F3%29x-3
Now your going to graph both lines but erase 1/2 of each line where the two lines cross. KEEP the blue line going to the right (for y=3x-10 if x>3) and only KEEP the part of the purple line that goes to the left ( for the x<= 3 condition.

Answer 13

so i am suppose to erase the bottom part of the blue where they meet and going back

Answer 14

ahhh

Answer 15

Correct, DON'T show the blue line left of where the two lines intercect. (Again where using the second graph, not the first).

Answer 16

You'll want to use two different colors or solid and dashed lines.

Answer 17

then what exactly am i doing with the purple line?

Answer 18

The purple line will be the rest of your graph left of the (2,-1) intersection point.
The net effect is the line looks "bent" like a shallow "v".

Answer 19

ahhhhhhh why lol jk

Answer 20

???

Answer 21

so i am not supose to do anything to the purple line?

Answer 22

no

Answer 23

ok ok i thank you guys for all your help!!!!!!!!!!!!......:)

Answer 24

Right - if you started out by graphing both lines - you would _erase_ the purple line to the RIGHT of (2,-1) --> but keep the purple line LEFT of (2,-1).
Do the same but in reverse for the blue line.
So looking at the graph - coming from the right. ONLY blue ray, SOLID purple dot at (2,-1) then ONLY purple ray to negative infinity.

Answer 25

np that is way i am here i am in 8 so if you need help

Answer 26

Sorry - lag in the reply posting.

Answer 27

lol its ok

Answer 28

I found a white board.
Go to this URL in another tab.
http://www.twiddla.com/491709

Answer 29

i am there

Answer 30

I didn't put in the x or y-axis.

Answer 31

ok i did that if u can see it on the board

Answer 32

Do you think you got it now
the function follows the blue line for all x inputs above 3 and the purple line for inputs 3 or less. The out-put "steps" over to another rule (function) at x=3.

Answer 33

So you set?

Answer 34

yes i do...and i thank u alot for ur help...:)

Answer 35

OK - good luck.